A battery consists of n rows of cells connect | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given: Total number of cells $N = mn = 24$. External resistance $R = 3 \, \Omega$. Internal resistance of each cell $r = 0.5 \, \Omega$.For maximu

Vedclass · Accepted Answer

$m = 12, n = 2$

Vedclass · Answer

(A) Given: Total number of cells $N = mn = 24$. External resistance $R = 3 \, \Omega$. Internal resistance of each cell $r = 0.5 \, \Omega$.For maximum current in a mixed grouping of cells,the external resistance must be equal to the total internal resistance of the battery.The total internal resistance of the battery is given by $R_{int} = \frac{mr}{n}$.Setting $R = R_{int}$,we get $3 = \frac{m 	imes 0.5}{n}$,which simplifies to $3 = \frac{m}{2n}$,or $m = 6n$.Substitute $m = 6n$ into the equation $mn = 24$:$(6n) 	imes n = 24 \implies 6n^2 = 24 \implies n^2 = 4 \implies n = 2$.Now,find $m$: $m = 6 	imes 2 = 12$.Thus,$m = 12$ and $n = 2$.

Similar Questions

$n$ identical cells,each of $e.m.f.$ $E$ and internal resistance $r$,are connected in series. This combination is connected in series with an external resistor $R$. What is the current flowing through $R$?

What is meant by the series connection of cells? Derive the equation for the equivalent electromotive force $(emf)$ of two cells with $emf$ $\varepsilon_1$ and $\varepsilon_2$ connected in series.

Ten identical cells each emf $2 \, V$ and internal resistance $1 \, \Omega$ are connected in series with two cells wrongly connected. $A$ resistor of $10 \, \Omega$ is connected to the combination. What is the current through the resistor (in $ \, A$)?

Four identical cells of emf $E$ and internal resistance $r$ are to be connected in series. If one of the cells is connected wrongly,the equivalent emf and effective internal resistance of the combination are

Vedclass Test Series

Exam Paper Generator

Online Exam Module